Question

Im trying to figure out how i can solve an equation that has been stored in an array.
I need some guidance on how to conquer such problem.
here is my conditions

• i have an array int X[30];
• and in there i have stored my desired equation: 5+6*20/4
• as well, i couldnt store the operants (+ / – * ) so i used different identifiers for them like ( -1 -2 -3 -4 ) because there should not be a negative value in the equation.

any help would be greatly appreciated

thanks

Answer 1

Normally, you have to define the priority of each operation and process each of them one by one.

First your expression is:        '`[ 5,-1, 6,-4,20,-2, 4]`'
Do all '/' first:                '`[ 5,-1, 6,-4, 5,-1, 0]`' <- 20/ 4 =  5+0
Then, do all '*':                '`[ 5,-1,30,-1, 0,-1, 0]`' <-  6* 5 = 30+0
Then, do all '-':                '`[ 5,-1,30,-1, 0,-1, 0]`' <-  Nothing
Then, do all '+' (sum positive): '`[35,-1, 0,-1, 0,-1, 0]`' <-  5+30+0+0 = 35 + 0 + 0

ADDED: Here is the C code.

void ShowArray(int pX[], int pLength) {
    int i;
    for(i = 0; i < pLength; i++)
        printf("%3d ", pX[i]);
    printf("\n");
}
 
void ShiftArray(int pX[], int pIndex, int pSkip, int pLength) {
    int i;
    for(i = pIndex; i < (pLength - pSkip); i++)
        pX[i] = pX[i + pSkip];
 
    for(i = (pLength - pSkip); i < pLength; i++)
        pX[i] = 0;
}
 
int main(void) {
    const int OPERCOUNT =  4;
    const int PLUS      = -1;  // -1 Do last
    const int SUBT      = -2;
    const int MULT      = -3;
    const int DIV       = -4;  // -4 Do first
 
    int X[]    = {5, PLUS, 6, MULT, 20, DIV, 4};
    int XCount = 7;
 
    ShowArray(X, XCount);
 
    int i;
    for(i = OPERCOUNT; --i >= 0; ) {
        int OPER = -(i + 1);

        int j = 0;
        for(j = 0; j < XCount; j++) {
            int x = X[j];
            if (x == OPER) {
                if      (x == PLUS) X[j - 1] = X[j - 1] + X[j + 1];
                else if (x == SUBT) X[j - 1] = X[j - 1] - X[j + 1];
                else if (x == MULT) X[j - 1] = X[j - 1] * X[j + 1];
                else if (x == DIV ) X[j - 1] = X[j - 1] / X[j + 1];
                ShiftArray(X, j, 2, XCount);
            }
        }
 
        ShowArray(X, XCount);
    }
 
    int Sum = 0;
    int j;
    for(j = 0; j < XCount; j++) {
        int x = X[j];
        if (x > 0) Sum += x;
    }
    printf("Result: %d\n", Sum);
}

And here is the result:

  5  -1   6  -3  20  -4   4 
  5  -1   6  -3   5   0   0 
  5  -1  30   0   0   0   0 
  5  -1  30   0   0   0   0 
 35   0   0   0   0   0   0 
Result: 35

This should works.

Enjoy coding.